Question: What is the area of the region between the graphs of $f(x)=-x^2+2x+12$ and $g(x)=x^2-12$ from $x=-3$ to $x=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{343}{3}$ (Choice B) B $\dfrac{52}{3}\sqrt{13}-1-32\sqrt{3}$ (Choice C) C $\dfrac{83}{3}$ (Choice D) D $7$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${2}$ ${4}$ ${\llap{-}2}$ ${5}$ ${10}$ ${\llap{-}10}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=-3$ and $x=4$. From this we are looking to evaluate: $ \int_{-3}^{4}\left( f(x)-g(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-3}^{4} \left( -x^2+2x+12- (x^2-12) \right) \,dx \\\\ &= \int_{-3}^{4} \left( -2x^2+2x+24 \right) \,dx\\\\ &= -\dfrac{2x^3}{3}+x^2+24x~\Bigg|_{-3}^{4} \\\\ &= \left( -\dfrac{128}{3}+16+96 \right) -\left( 18+9-72 \right)\\\\ &= \dfrac{343}{3} \end{aligned}$ Answer The area is $\dfrac{343}{3}$ square units.